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Nutka1998 [239]
2 years ago
5

Solve 10e^2x - 5 =23^x for x.

Mathematics
2 answers:
yawa3891 [41]2 years ago
5 0

10e^{2x}-5=23e^x\qquad\text{subtract}\ 23e^x\ \text{from both sides}\\\\10e^{2x}-23e^x-5=0\\\\10(e^x)^2-23(e^x)-5=0\\\\\text{substitution:}\ e^x=t > 0\\\\10t^2-23t-5=0\\\\10t^2-25t+2t-5=0\\\\5t(2t-5)+1(2t-5)=0\\\\(2t-5)(5t+1)=0\iff2t-5=0\ \vee\ 5t+1=0\\\\2t=5\ \vee\ 5t=-1\\\\t=\dfrac{5}{2} > 0\ \vee\ t=-\dfrac{1}{5} < 0\\\\\text{therefore}\ e^x=\dfrac{5}{2}\to\ln e^x=\ln\left(\dfrac{5}{2}\right)\\\\\boxed{x=\ln\left(\dfrac{5}{2}\right)}

bixtya [17]2 years ago
5 0

Answer:

B is the correct option choice !


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A triangle is given to us. In which one angle is 30° and length of one side is 18ft ( hypontenuse) .So here we can use trignometric Ratios to find values of rest sides. Let's lable the figure as ∆ABC .

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